Let $X$ be a $T_1$ topological space. Show that $X$ is connected and locally connected if and only if for each open cover $\{U_s\}_{s\in S}$ and any $x, y \in X$, there exist $s_1,. . . , s_n \in S$ and $V_1,. . . , V_n$ connected open in $X$ such that $x \in V_1$, $y \in V_n$, $V_i \subset U_{s_i}$ for all $i \in \{1,. . . , n\}$ and $V_i \cap V_j \neq \emptyset$ if, and only if, $| i - j | \leq 1$.
My idea: Let's suppose for each open cover $\{U_s\}_{\{s\in S\}}$ and any $x, y \in X$, there exist $s_1,. . . , s_n \in S$ and $V_1,. . . , V_n$ connected open in $X$ such that $x \in V_1$, $y \in V_n$, $V_i \subset U_{s_i}$ for all $i \in \{1,. . . , n\}$ and $V_i \cap V_j \neq \emptyset$ if, and only if, $| i - j | \leq 1$ and let's see that $ X $ is connected and locally connected
- Connected. Suppose $X$ is disconnected; that is, $X = U_1 \cap U_2$, where $U_1 \cap U_2 = \emptyset$, $U_1$, $U_2$ open not empty. Notice that $\mathcal{U}= \{U_1, U_2\}$ is an open cover of $X$. Lets $x \in U_1$ and $y \in U_2$, by hypothesis there are $s_1,s_2 \in \{1,2\}$ and $V_1, V_2$ connected open in $X$ such that $x \in V_1$, $y \in V_2$, $V_i \subset U_{s_i}$ for all $i \in \{1, 2\}$, since U and V are disjoint, it contradicts that $V_1 \cap V_2 \neq \emptyset$. therefore $ X $ is connected.
- locally connected. Let $x \in X$, and $U$ an open of $ X $ such that $x \in U \subset X$, let $V$ be a closed neighborhood of $x$ in $X$ such that $V \subset U$ consider cover $\mathcal=\{U, X \setminus V\}$. let $x \in U$, $y \in X \setminus U$ by hypothesis there are $V_1$ open connected such that $x \in V_1 \subset U$. I am not sure if this last argument is correct.
The other implication I'm stuck, some help please
